@Bhargav: That's what LaTeX always does. For in-line formulas, use a single $.
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Arturo MagidinAug 14 '11 at 2:30

6

@Bhargav: This isn't so much about an infinite matrix as it is about an infinite series with double indices. The fact that it's a matrix is irrelevant. Just take $A_{mn} = \frac{m}{m+n}$.
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Arturo MagidinAug 14 '11 at 2:31

Ya now i get it , rather than looking at it fromt he point of a matrix question ,i should have seen t as a calculus quetsion then the problem would have been solved. Thx arturo and zev
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BhargavAug 14 '11 at 2:37

2 Answers
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$$\lim_{n\to\infty}A_{mn}=\lim_{n\to\infty}\left(\frac{m}{m+1}\right)^n=0$$
because $0<\frac{m}{m+1}<1$ for all $m\geq 1$, and
$$\lim_{m\to\infty}A_{mn}=\lim_{m\to\infty}\left(\frac{m}{m+1}\right)^n=\left(\lim_{m\to\infty}\frac{m}{m+1}\right)^n=1^n=1.$$

wow, you don't even have to know much about limits for this!
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GEdgarAug 14 '11 at 13:05

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Exercise for the OP: you can even modify Jonas's answer to get a yet different limit for all "diagonals", that is, $\lim_{m \to \infty} A_{m,m+k} = \lim_{n\to\infty} A_{n+k,n} = \frac12$ for $k\geq 0$
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Willie WongAug 14 '11 at 13:51